A Proof of thermodynamic stability condition

AI Thread Summary
The discussion centers on understanding a mathematical detail in the proof of the thermodynamic stability condition from Kardar's Statistical Mechanics course. The author grapples with the relationship between intensive and extensive variables in a homogeneous system at equilibrium, particularly when dividing the system into two subsystems that exchange energy. A key point is the clarification that, at equilibrium, the intensive variables of both subsystems (temperature, energy density, and chemical potential) must be equal, which resolves the confusion regarding their variations. The conversation highlights the importance of recognizing that while the subsystems may exchange energy, their intensive variables remain consistent due to the overall equilibrium condition. The dense nature of Kardar's text is acknowledged as a challenge for comprehension.
chimay
Messages
81
Reaction score
8
TL;DR Summary
Question about a mathematical detail in the proof of the thermodynamic stability relation
I am watching Kardar's Statistical Mechanics course in my spare time and I am struggling to understand a mathematical detail in the proof of the thermodynamic stability condition. See Eq. I.62 here.

The author considers a homogeneous system at equilibrium with intensive and extensive variables (T,J,\mu) and (E,x,N), respectively. Then, He imagines that the latter system is divided into two subsystems (A and B) that can exhange energy. Under the assumption that E,x and N are conserved, we have \delta E_A=-\delta E_B, \delta x_A=-\delta x_B and \delta N_A=-\delta N_B. What is not clear to me is the statement "Since the intensive variables are themselves functions of the extensive coordinates, to first order in the variations of (E, x, N), we have \delta T_A=-\delta T_B, \delta J_A=-\delta J_B and \delta \mu_A=-\delta \mu_B."

Can anyone explain to me the previous statement more in detail?

Thank you!
 
Science news on Phys.org
If, for example, ##T=f(E)##, then to the first order, ##\delta T=f'\, \delta E##.
Then, ##\delta E_A = -\delta E_B## implies ##\delta T_A = -\delta T_B##.
 
Hi Hill, thank you for your reply.
I understand your point:
\delta T_A = \frac{\partial f}{\partial E}\bigg\rvert_{T_A} \delta E_A and
\delta T_B = \frac{\partial f}{\partial E}\bigg\rvert_{T_B} \delta E_B.

However, since the two subsystems are not necessarily equal, even though \delta E_A = - \delta E_B, \delta T_A \ne \delta T_B because \frac{\partial f}{\partial E}\bigg\rvert_{T_A} \ne \frac{\partial f}{\partial E}\bigg\rvert_{T_B}.

Where am I wrong?
 
chimay said:
However, since the two subsystems are not necessarily equal, even though δEA=−δEB, δTA≠δTB because ∂f∂E|TA≠∂f∂E|TB.
They say, "The two subsystems, A and B, initially have the same values for the intensive variables". That is, ##T_A=T_B##. Then, ##\frac{\partial f}{\partial E}\bigg\rvert_{T_A} = \frac{\partial f}{\partial E}\bigg\rvert_{T_B}##.
 
You are right! I read it so many times but I missed that. Indeed, since the overall system is at equilibrium, it makes sense that the two "subsystems" have the same values for the intensive variables. Kardar's book is very dense of information and I find it hard to follow sometimes...

Thank you for your help!
 

Similar threads

Back
Top